Problem: Evaluate \[\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}},\]giving your answer as a fraction in lowest terms.
Using the property $\log_a b^x = x \log_a b,$ we have \[\begin{aligned} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} &= \frac{2}{6\log_4 2000} + \frac{3}{6\log_5 2000} \\ &= \frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000}. \end{aligned}\]Since $\log_a b = \frac1{\log_b a}$, we can then write \[\frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000} = \frac{1}{3}\log_{2000} 4 + \frac{1}{2}\log_{2000} 5,\]which equals \[\log_{2000} (4^{1/3} 5^{1/2})= \log_{2000} (2^{2/3} 5^{1/2}).\]Since $2000 = 2^4 5^3 = \left(2^{2/3} 5^{1/2}\right)^6$, the expression $\boxed{\tfrac{1}{6}}$.